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Determining whether the series is convergent or divergent

  • Thread starter umzung
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  • #1
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Homework Statement


Determine if the series is convergent.

Homework Equations



∑ (((2n^2 + 1)^2)*4^n)/(2(n!))
n=1[/B]

The Attempt at a Solution


I'n using the Ratio Test and have got as far as (4*(2(n+1)^2+1)^2)/((n+1)((2n^2+1)^2)). I know this series converges but I need to find the limit to be < 1 to show this. Is there way to now divide each term by the dominant term n^2, or do I need to multiply the whole thing out and divide by the new dominant term? I've tried that and have found the limit to be 0/4 = 0.
 

Answers and Replies

  • #2
Orodruin
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Just note that the numerator goes as ##8n^4 + \mathscr O(n^3)## and that the denominator goes as ##4n^5 + \mathscr O(n^4)## as ##n\to \infty##. This will be sufficient.
 
  • #3
vela
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Homework Statement


Determine if the series is convergent.

Homework Equations



∑ (((2n^2 + 1)^2)*4^n)/(2(n!))
n=1[/B]

The Attempt at a Solution


I'n using the Ratio Test and have got as far as (4*(2(n+1)^2+1)^2)/((n+1)((2n^2+1)^2)). I know this series converges but I need to find the limit to be < 1 to show this. Is there way to now divide each term by the dominant term n^2, or do I need to multiply the whole thing out and divide by the new dominant term? I've tried that and have found the limit to be 0/4 = 0.
If you want to show the algebra, you can do something like this:
$$\frac{4}{n+1}\frac{[2(n+1)^2+1]^2}{(2n^2+1)^2}
=\frac{4}{n+1}\frac{\left[(n+1)^2\left(2+\frac{1}{(n+1)^2}\right)\right]^2}{\left[n^2\left(2+\frac{1}{n^2}\right)\right]^2}$$ so you don't have to multiply everything out.
 

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